Question: Simplify; express your answer in exponential form. Assume $n\neq 0, a\neq 0$. $\dfrac{{(n^{-2})^{-4}}}{{(n^{3}a^{2})^{-1}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{-2}}$ to the exponent ${-4}$ . Now ${-2 \times -4 = 8}$ , so ${(n^{-2})^{-4} = n^{8}}$ In the denominator, we can use the distributive property of exponents. ${(n^{3}a^{2})^{-1} = (n^{3})^{-1}(a^{2})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{-2})^{-4}}}{{(n^{3}a^{2})^{-1}}} = \dfrac{{n^{8}}}{{n^{-3}a^{-2}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{8}}}{{n^{-3}a^{-2}}} = \dfrac{{n^{8}}}{{n^{-3}}} \cdot \dfrac{{1}}{{a^{-2}}} = n^{{8} - {(-3)}} \cdot a^{- {(-2)}} = n^{11}a^{2}$.